7. Orange juice concentrate of density 1040 kg.m -3 and 3.5 mPa.s at a rate of 2
ID: 2995004 • Letter: 7
Question
7. Orange juice concentrate of density 1040 kg.m-3 and 3.5 mPa.s at a rate of 24 tonnes per hour through a pipe section that consists of at the inner surfaces of the pipes are hydraulically smooth. There is a straight horizontal pipe of internal diameter 4 cm of length 8 m upstream of a fully open gate valve and a further 2 m run of pipe. There is then a sudden expansion such that the flow area increases five-fold, and another 6 m run of horizontal pipe that has the expanded diameter. Estimate the pressure loss through this piping arrangement.
8. A domestic hot-water circulation pump running at 2850 rpm has the following performance:
Q (m3/h)
0
1
2
3
H (m)
2.7
2.3
1.8
1.2
The pump is used to circulate water at 90
Q (m3/h)
0
1
2
3
H (m)
2.7
2.3
1.8
1.2
Explanation / Answer
1.Q = rho*V*D
24000 = 1040*V*0.04
V = 576.92 = 577 m/hr = 0.16 m/sec
Re = rho*V*D/u = 1040*0.16*0.04/ 0.0035 = 1901.7 < 2100 So, laminar flow.
From Hagen Poiseulle equation, deltaP = 32*L*u*V/(D^2) where u is viscosity.
Since, for the fully open gate valve we can neglect the pressure drop across it.
Pressure drop across first (8+2) = 10m
deltaP1 = 32*10*0.0035*0.16/0.04^2 = 112 Pa
Across expansion, from continuity equation, A1V1 = A2V2 where A2 = 5*A1
So, V2 = 0.16/5 = 0.032
From Bernoulli equation, energy conservation implies,
deltaP2 = 0.5*rho*(V1^2 - V2^2) = 12.78Pa
deltaP3 = 32*6*0.0035*0.032/(5*(0.04^2)) =2.69
So, total pressure drop = 112+12.78+2.69 = 127.47 Pa.
2. Throughinterpolation Q = 1.5 m3/h, H = 2.05.
Therefore, pressure drop = rho*g*H = 965*9.81*2.05 = 19406.63 Pa
Therefore, d^2 = 32*50*2.14*10^-5 * V/ deltaP where Q = rho*V*S where S = pi*d^2
d^4 = 32*50*2.14*10^-5 * 1.5/ (19406.63*965*pi)
d = 5.4mm.
we know that N1/N2 = Q1/Q2 So, Q2 = 1500*2000/2850 =1052.63 lt/hr
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